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Your Differentiability Toolkit: All Conditions, Theorems & Pitfalls

Everything you need to master differentiability for JEE Main & Advanced in one comprehensive guide.

7
Key Theorems
5
Essential Conditions
8
Common Pitfalls
100%
JEE Coverage

Why Differentiability Matters in JEE

Differentiability is the foundation of calculus and appears in every JEE mathematics paper. Understanding the precise conditions and common pitfalls can help you secure 4-8 marks consistently.

🎯 JEE Exam Pattern Insight

  • 1-2 questions directly test differentiability conditions
  • 3-4 questions use differentiability as a stepping stone
  • Questions often combine continuity + differentiability
  • Common formats: MCQs, numerical value, and comprehension

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Basic Conditions Key Theorems Common Pitfalls Practice

1. Fundamental Conditions for Differentiability

Condition 1: Continuity is Necessary

Theorem Statement

If a function $f(x)$ is differentiable at $x = a$, then it must be continuous at $x = a$.

$$ \text{Differentiable} \Rightarrow \text{Continuous} $$

But the converse is not necessarily true!

📚 Example: $f(x) = |x|$ at $x = 0$

Check continuity: $\lim_{x \to 0} |x| = 0 = f(0)$ ✓ Continuous

Check differentiability:

Left derivative: $\lim_{h \to 0^-} \frac{|h|}{h} = -1$

Right derivative: $\lim_{h \to 0^+} \frac{|h|}{h} = 1$

Result: LHD ≠ RHD ⇒ Not differentiable at x = 0

Condition 2: Left & Right Derivatives Must Equal

Mathematical Definition

A function $f(x)$ is differentiable at $x = a$ if:

$$ \lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h} = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h} $$

And this common value is finite.

⚠️ Common Mistake

Students check only one-sided limit or forget to verify both limits exist and are equal.

Condition 3: Smoothness Requirement

Geometric Interpretation

A function is differentiable at a point if its graph has a unique tangent at that point with finite slope.

No sharp corners, vertical tangents, or discontinuities.

📚 Non-Differentiable Cases

Sharp Corner:
$f(x) = |x|$ at x = 0
Vertical Tangent:
$f(x) = x^{1/3}$ at x = 0
Cusp:
$f(x) = x^{2/3}$ at x = 0

2. Essential Theorems for JEE

Theorem 1: Chain Rule

Statement

If $y = f(u)$ and $u = g(x)$ are both differentiable, then:

$$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $$

📚 JEE Application

Find derivative of $y = \sin(x^2 + 1)$

Let $u = x^2 + 1$, then $y = \sin u$

$\frac{dy}{du} = \cos u$, $\frac{du}{dx} = 2x$

$\frac{dy}{dx} = \cos(x^2 + 1) \cdot 2x$

Theorem 2: Product Rule

Statement

If $u(x)$ and $v(x)$ are differentiable, then:

$$ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) $$

⚠️ Common Error

Students often write $(uv)' = u'v'$ which is completely wrong!

Theorem 3: Rolle's Theorem

Conditions & Conclusion

If $f(x)$ is:

  1. Continuous on $[a, b]$
  2. Differentiable on $(a, b)$
  3. $f(a) = f(b)$

Then ∃ at least one $c \in (a, b)$ such that $f'(c) = 0$.

📚 Application Example

Verify Rolle's theorem for $f(x) = x^2 - 4x + 3$ on $[1, 3]$

$f(1) = 0$, $f(3) = 0$ ✓ Condition 3 satisfied

$f'(x) = 2x - 4 = 0 \Rightarrow x = 2$

$c = 2 \in (1, 3)$ where $f'(2) = 0$ ✓

Theorem 4: Mean Value Theorem

Statement

If $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then ∃ $c \in (a, b)$ such that:

$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

⚠️ Common Confusion

Students think MVT guarantees exactly one 'c', but it only guarantees at least one.

3. Common Pitfalls & How to Avoid Them

Pitfall 1: Assuming Continuity Implies Differentiability

The Error

Thinking: "Function is continuous, so it must be differentiable."

📚 Counterexample

$f(x) = |x - 1|$ is continuous everywhere but not differentiable at $x = 1$.

✅ Prevention Strategy

Always check both:

  • Continuity at the point
  • Equality of left and right derivatives

Pitfall 2: Misapplying Chain Rule

The Error

Forgetting to multiply by the derivative of the inner function.

📚 Wrong vs Right

Wrong:
$\frac{d}{dx}[\sin(x^2)] = \cos(x^2)$
Correct:
$\frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x$

Pitfall 3: Forgetting Domain Restrictions

The Error

Differentiating functions at points where they're not defined or not differentiable.

📚 Critical Points

  • $f(x) = \sqrt{x}$ at $x = 0$ (vertical tangent)
  • $f(x) = \frac{1}{x}$ at $x = 0$ (not defined)
  • $f(x) = |x-2|$ at $x = 2$ (corner)

📋 Differentiability Checklist

Use this checklist when solving differentiability problems:

4. Practice Problems

Problem 1: Piecewise Function

Check differentiability of $f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 2x-1 & \text{if } x > 1 \end{cases}$ at $x = 1$

Problem 2: Absolute Value Function

Find all points where $f(x) = |x^2 - 4|$ is not differentiable.

Key Takeaways

🎯 Must Remember

  • Differentiable ⇒ Continuous (but not vice versa)
  • Always check both left and right derivatives
  • Sharp corners, cusps, vertical tangents break differentiability
  • Chain rule: don't forget the inner derivative

🚀 Exam Strategy

  • Use the 6-step checklist systematically
  • Practice piecewise functions thoroughly
  • Memorize common non-differentiable cases
  • Time yourself on differentiability problems

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